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Comments:
<0> ARGHHHHHH
<0> that's what I got why doesn't it work?
<1> Alright, the obvious answer is the right answer
<1> Thanks!
<2> SirJective, really?
<0> okay, how do I make a function in mathematica?
<3> Calm down, Squall`
<0> the function F(x)?
<3> That answer is the same as yours
<0> Yeah, I know.
<4> well, by an algebraic definition of gcd, not by a definition using the usual order on the naturals
<0> But I keep getting the wrong indefinite integral.
<0> *definite integral
<3> Ah
<0> following that, I should get -2 ln 4 + 2 ln 2 + 2 ln 3, b ut mathematica says that's wrong
<2> Ahh, right, because 0 divides 0, and any number that divides 0 must divide 0
<4> kilimanjaro: exactly. :-)
<5> the standard topology on Z is the discrete one right?
<4> toed: for most uses, yes.
<0> % N[-2 Log[4] + 2 Log[2] + 2 Log[3]]
<6> Squall`: 0.8109302162163288
<5> excellent
<0> % N[Integrate[-2/(x + 1) + 2/(x - 1), x, {2, 3}]]
<6> Squall`:
<6> Integrate::ilim: Invalid integration variable or limit(s) in {2, 3}.
<6> Integrate[2/(-1 + x) - 2/(1 + x), x, {2, 3}]
<7> I have some gl*** with water. I need some function of height of water respect to coordinate x and inclination of the gl*** teta... can some one help me to obtain it???
<0> % N[Integrate[-2/(x + 1) + 2/(x - 1), {x, 2, 3}]]
<6> Squall`: 0.8109302162163288
<0> wait, wtf...
<0> I did this right the first time?
<0> % Integrate[-2/(x + 1) + 2/(x - 1), {x, 2, 3}]
<6> Squall`: Log[9/4]
<0> % Integrate[1 / (x^2 - 1), {x, 2, 3}]
<6> Squall`: Log[3/2]/2
<0> Please kill me....
<0> N[.5 Log[1.5]]
<0> % N[.5 Log[1.5]]
<6> Squall`: 0.2027325540540822
<0> Um... is .2027 = .8109 by any chance?
<0> *meltdown*
<7> % x
<8> Squall`: no, I don't think so.
<6> webito: x
<0> I didn't do anything wrong.... :( I've been doing all my homework, and everything. Integrals hate me
<0> I'm about to sob manly tears.
<8> If you find a contradiction in mathematics, turn it into a proof of all the Millenium Prize Problems.
<0> I just did. .2027 is actually equal to .8109 All math is wrong
<2> Squall`, what is the problem?
<2> integrate dx/(x^2 - 1) over [2,3]?
<0> I want to find out why Integrate[-2/(x + 1) + 2/(x - 1), {x, 2, 3}] = 1/2 ln(3/2) instead of ln (9/4)
<0> yes
<0> I used partial fractions and got -2/(x+1) + 2/(x-1)
<2> Squall`, umm
<0> but look at this:
<0> % Integrate[-2/(x+1), 2/(x-1), {x, 2, 3}]
<6> Squall`:
<6> 2
<6> Integrate::ilim: Invalid integration variable or limit(s) in ------.
<6> -1 + x
<6> Integrate[-2/(1 + x), 2/(-1 + x), {x, 2, 3}]
<0> sorry about that bad equation
<3> Ehm
<0> didn't mean to spam the room
<3> You have the 2 upside down
<0> where?
<3> 1/(x^2-1) = -1/2(x+1) + 1/2(x-1)
<0> That means that the form of 1 / (x^2 - 1) is 1/A(x+1) + 1/B(x - 1), right?
<0> 16:32] Squall`: Wait, so if you have 1 / (x^2 - 1), the form of the partial fraction is 1/A(x-1) + 1/B(x+1) instead of A and B being in the numerator?
<0> [16:33] me22|mourning: Squall` : doesn't make a difference
<0> omg it does make a difference, doesn't it?
<3> It doesn't make a difference
<0> hahahahahahaha
<3> If you put them in the denominator, you'll just get the reciprocal of what you'd get if you put it in the numerator
<3> Either way it'll come out right if you do the algebra right
<0> Then why isn't the answer also 2/(x+1) - 2/(x - 1)?
<0> because I got
<0> A / (x + 1) + B / (x - 1), where 1 = A(x - 1) + B(x + 1)
<0> if you make x = 1, B = 2
<3> No...
<0> Please for the love of god tell me I'm wrong.
<0> yes
<3> You get 1 = 2B
<0> how?
<0> wait
<0> oh my god...
<0> Thank you so much.
<0> I love you. I'm seriously in love with you. I don't care if you're male, female, human, dolphin, whatever, I love you.
<0> If I wasn't so stubborn in find the answer to the current problem even if it's taking me hours before doing the rest, I'd probably have a lot more free time =X
<3> Good, because I have a sum of $2000000000 which I need to transfer from country of Nigeria and perhaps you could send me bank account number for temporary transfer
<0> Totally. I will do that for you, I'll do anything. $2,000,000,000.00 <--- look, I even formatted it for you
<9> Kasadkad: sure thing, i'll just pile it all into an e-gold account or something
<0> What? I have to use the quadratic formula in an Integration problem? What is this, Algebra I?
<10> Squall`: yes, Algebra is used forever. Like most math, you can't forget what you learned last [whatever], since you'll use it again.
<0> Pfff, I ain't doing no quadratic formula. Time to crank out the calculator
<10> Squall`: do you lament having to do addition as well, because you already p***ed second grade?
<0> yes
<0> I try to avoid it when I can
<11> heh, I never bother with quadratic formula
<11> just complete the square manually, most of the time :P
<0> I... don't know how to complete the square. i don't think I was awake when I learned that in middle school. I guess I could probably learn it in 1 or 2 minutes
<0> I don't even know what that means
<11> turning x^2+bx+c into (x-h)^2+k
<11> quadratic formula falls out from it quite trivially
<8> I never complete the square or factor; I just use the quadratic formula.
<0> lol
<8> I prefer mathematical formulas to mathematical methods.
<8> Well, okay, I do complete the square when putting an ellipse into whatever standard form.
<12> ( Squall` ] Totally. I will do that for you, I'll do anything. $2,000,000,000.00 <--- look, I even formatted it for you < incorrectly formatted =(
<0> Kinks: What country do you live in?
<12> uhm, canada.
<0> So, how is that formatted wrong?
<12> when you have 10 thousand it isn't 1,000,0
<12> it's 10,000
<3> It's formatted fine
<12> if you say so..
<0> uh, what?
<12> what you did was add a comma after the first nonzero number and add another one for every 3 0s
<12> which I've never once seen done before and don't see how it makes any sense whatsoever
<12> "normally" commas are added once every 3 digits going left starting from the decimal
<12> but who knows, maybe I'm wrong.
<12> I really don't care to make a big deal out of it
<0> commas don't start at the decimal
<11> $2,000,000,000.00 is american
<11> $2 000 000 000,00 is used in some places
<11> 2_000_000_000.00 in perl
<0> Doesn't Canada use the same locale as the US?
<12> oh
<12> I was seeing it wrong
<12> I thought the . was a comma.
<12> so my fault.
<12> given that I hope you see where my idea of "incorrect formatting" came from.
<0> yeah
<0> Hm... another partial fractions question, I have Integral of 10/(x-1)(x^2 + 9)dx, so I factored a 10 out to the left of the integral, leaving me (x-1)(x^2 + 9)
<0> Which takes the form A/(x-1) + (Bx +C)/(x^+9)
<0> Which takes the form A/(x-1) + (Bx +C)/(x^2+9)
<0> so, I got 1 = A(x^2 + 9) + (Bx + C)(x-1)
<0> Supposed we let x = 1, , A = 1/10
<0> So we have 10 * Integral((1/10(x-1) + (Bx +C)/(x^2 + 9))dx)
<0> the first term integrates to ln | x - 1| and due to some weird formula in my book, (Bx + C)/(x^2 + 9) = 10/3Arctan(x/3)
<0> but the antiderivative has 3 terms
<0> I'm not sure where the middle term comes from.
<0> well, 4 if you include the arbitrary constant of integration
<13> hello
<13> EXOTICS ADULT FORUM :::::::::: http://exotics.ezbbforum.com
<13> EXOTICS ADULT FORUM :::::::::: http://exotics.ezbbforum.com ...
<13> EXOTICS ADULT FORUM :::::::::: http://exotics.ezbbforum.com
<14> ...
<15> hey just what #math needs
<16> hi
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